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Math Help - trigonometric

  1. #1
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    trigonometric


    I have 2 questions:

    1. find for z ; where z^3=sqrt[imaginary]
    2. Simplify: sin x/(1-cos x)
    can somebody help me with these 2 questions showing working out?
    Please & Thank You
    Last edited by mr fantastic; March 17th 2009 at 02:53 AM. Reason: Deleted overuse of smilies.
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  2. #2
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    Hello, n7167113!

    Find z\!:\;\;z^3\:=\:\sqrt{i}
    We have: . z \:=\:i^{\frac{1}{6}}

    We know: . i \:=\:\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2},

    Use DeMoivre's Formula for the six roots of: . \left(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}\right)^{\frac{1}{6}}




    Simplify: . \frac{\sin x}{1-\cos x}
    This doesn't simplify very much, but here goes . . .


    Multiply by \frac{1 +\cos x}{1+\cos x}\!:\;\;\frac{\sin x}{1 - \cos x}\cdot\frac{1+\cos x}{1+\cos x} \;=\;\frac{\sin x(1 + \cos x)}{1-\cos^2\!x}  \;=\;\frac{\sin x(1 + \cos x)}{\sin^2\!x}

    . . = \;\frac{1+\cos x}{\sin x} \;=\;\frac{1}{\sin x} + \frac{\cos x}{\sin x} \;=\;\csc x + \cot x

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  3. #3
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    For the simplification one, try multiplying both the numerator and denominator by (1+cos(x)), and from there you should be able to get a nice answer

    edit> I was beat to the punch
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